What is (\partial)/(\partial x)(3x^3+4x^2y-4y^3) ?

The answer to (\partial)/(\partial x)(3x^3+4x^2y-4y^3) is 9x^2+8xy

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(\partial)/(\partial x)(3x^3+4x^2y-4y^3)

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - 4 y^{3})

9 x^{2} + 8 x y

Solution steps

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - 4 y^{3})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (3 x^{3}) + \frac{\partial}{\partial x} (4 x^{2} y) - \frac{\partial}{\partial x} (4 y^{3})

\frac{\partial}{\partial x} (3 x^{3}) = 9 x^{2}

\frac{\partial}{\partial x} (4 x^{2} y) = 8 y x

\frac{\partial}{\partial x} (4 y^{3}) = 0

= 9 x^{2} + 8 y x - 0

Simplify

= 9 x^{2} + 8 x y

\frac{d}{d x} (5 ln (sec (x) + tan (x)))

\int \frac{( arctan ( x ) ) ^{2}}{1 + x ^{2}} d x

y y^{^{'}} + x y^{2} = (e^{- x^{2}})

f (x) = \frac{1}{3} x^{3} + 5.5 x^{2} + 30 x + 22

x \to 0 + lim (\frac{x ^{x} - 1}{2 x})

What is (\partial)/(\partial x)(3x^3+4x^2y-4y^3) ?
The answer to (\partial)/(\partial x)(3x^3+4x^2y-4y^3) is 9x^2+8xy